Exercise 9.A.17

Proof. (a) Obviously $V$ closed under addition and complex scalar multiplication. One can easily check that the other properties in 1.19 are satisfied.

(b) Let $n$ be the integer such that $\dim <!--\; FUNCTION\; APPLICATION\; -->V=2n$ and consider the following process.

• Step $1$
  Choose a nonzero $v_1\in V$. Then $v_1,T{v}_1$ is linearly independent in $V$ as real vector space because $v_1$ is not an eigenvector of $T$ (because $T$ has no eigenvectors). Set $U_1=\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,T{v}_1)$. Then $\dim <!--\; FUNCTION\; APPLICATION\; -->U_1=2$ and $U_1$ is invariant under $T$.

• Step $j$
  

  If $j=n+1$, stop the process. We have that

  $$\dim <!--\; FUNCTION\; APPLICATION\; -->U_{j-1}=2(j-1)\le 2n-2<\dim <!--\; FUNCTION\; APPLICATION\; -->V,$$

  $U_{j-1}$ is invariant under $T$ and

  $$v_1,T{v}_1,\dots ,v_{j-1},T{v}_{j-1}$$

  is a basis of $U_{j-1}$. Hence there exists a nonzero $w\in V$ such that $w\notin U_{j-1}$. Since $T$ is surjective, there exists $v_j\in V$ such that $T{v}_j=w$. Thus $v_j\notin U_{j-1}$, because $U_{j-1}$ being invariant under $T$ would imply $w\in U_{j-1}$. Moreover, the list

  $$v_1,T{v}_1,\dots ,v_j,T{v}_j$$

  is linearly independent. To see this, let $a_1,\dots ,a_j,c_1,\dots ,c_j\in ℝ$ such that

  $$a_1{v}_1+c_{1}T{v}_1+\cdots +a_j{v}_j+c_{j}T{v}_j=0$$

  We already know that $v_j$ is not in the span of the previous vectors, so $c_j=0$ implies $a_j=0$ which implies that the rest of the $a$’s and $c$’s is $0$. Assume by contradiction $c_j\ne 0$. We can write the equation above as

  $$a_1{v}_1+c_{1}T{v}_1+\cdots +a_{j-1}{v}_{j-1}+c_{j-1}T{v}_{j-1}=-a_j{v}_j-c_{j}T{v}_j.$$

  Applying $T$ to both sides we get

  $$a_{1}T{v}_1-c_1{v}_1+\cdots +a_{j-1}T{v}_{j-1}-c_{j-1}{v}_{j-1}=-a_{j}T{v}_j+c_j{v}_j.$$

  Multiplying $(3)$ by $\frac{a_j}{c_j}$ and summing with $(2)$ shows us that

  $$\left(\frac{-a_j^2}{c_j}-c_j\right)T{v}_j\in U_{j-1}.$$

  Hence the parentheses above equals $0$. This means that $-a_j^2=c_j^2$, which is only possible if $a_j=c_j=0$ and contradicts our assumption that $c_j\ne 0$. Therefore the list in $(1)$ is linearly independent. Set

  $$U_j=\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,T{v}_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_j,T{v}_j).$$

  Then $\dim <!--\; FUNCTION\; APPLICATION\; -->U_j=2j$ and $U_j$ is invariant under $T$.

At the end of the process, we will have constructed a subspace $U_n$ of $V$ which has a basis

and dimension $2n$. Thus $U_n=V$ and the above list is a basis of $V$. Clearly $v_1,\dots ,v_n$ spans $V$ as a complex vector space. Furthermore, it is linearly independent in $V$ as a complex vector space, because if a (complex) linear combination of it equals $0$, then a linear combination of the list above equals $0$, which implies that the coefficients are $0$. Therefore $V$ as a complex vector space has dimension $n$, completing the proof. □